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A Path integral formula for a state with non-trivial time dependency

  1. Feb 3, 2017 #1

    ShayanJ

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    Consider a system with a time-dependent Hamiltonian. We know that the evolution of the state of this system, is given by ## \displaystyle |\Psi(t_1)\rangle=T \exp\left( -i \int_{t_0}^{t_1} dt H(t) \right) |\Psi(t_0)\rangle ##.
    Do you think you can prove that the path integral formula for the state(the one which was the ground state at ## t=-\infty ##) of this system, can be given by the formula below?

    ##\displaystyle \Psi(x,t)\propto \int_{z(t=-\infty)=const}^{z(t)=x} \mathcal D [z(t)] \ e^{iS[z(t)]} ##

    Thanks
     
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  3. Feb 5, 2017 #2

    stevendaryl

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    Nobody else stepped up to answer this, so I'll take a stab at it.

    Even though I haven't seen path integrals applied to time-dependent Hamiltonians, it seems to me that the derivation (such as it is---there is a big problem with convergence in the usual derivation) isn't affected by time-dependence.
     
  4. Feb 6, 2017 #3

    ShayanJ

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    The usual derivation is like below:
    ## \langle x|y,t \rangle=\langle x|e^{iHt}|y\rangle=\sum_b \langle x|e^{iHt}|\psi_b\rangle\langle \psi_b|y\rangle=\sum_b e^{iE_bt}\langle x|\psi_b\rangle\langle \psi_b|y\rangle=\sum_b e^{iE_bt}\psi_b(x) \bar \psi_b(y) ##.
    Now by a Wick rotation and taking the limit ## t_E\to -\infty ##, we'll have ## \psi_0(x) \propto \langle x|y,t_E=-\infty\rangle ##.
    But we know that ## \displaystyle \langle x|y,t\rangle\propto \int_{z(0)=x}^{z(t)=y} \mathcal D[z(t)] \ e^{iS[z(t)]} ##, so we have:

    ##\displaystyle \psi_0(x) \propto \int_{z(t_E=-\infty)=const}^{z(0)=x} \mathcal D[z(t)] \ e^{iS[z(t)]} ##.

    But for a time-dependent Hamiltonian, you can't separate the time-dependent Schrodinger equation, so the decomposition ## \Psi_b(x,t)=e^{-iE_bt}\psi_b(x) ## doesn't make sense. So ## \psi_0(x) ## doesn't have any meaning and also some steps in the proof depend on this decomposition.
     
  5. Feb 6, 2017 #4

    stevendaryl

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    Well, in the heuristic (non-rigorous) derivation that I first saw, there was no mention of decomposition into energy eigenstates. Also, I think we discussed on another thread why you use [itex]\langle x |y,t\rangle = \langle x| e^{+iHt}|y\rangle[/itex], but I don't remember the reason. The sources I've seen use [itex]\langle x |y,t\rangle = \langle x| e^{-iHt}|y\rangle[/itex].

    The basis that I've seen for developing path integrals starts with the existence of a Green's function [itex]G(x,t,x_0,t_0)[/itex] with the following properties:
    1. If [itex]\psi(x,t)[/itex] is a solution to the time-dependent Schrodinger equation [itex]H \psi = i \frac{d}{dt} \psi[/itex], then if [itex]t > t_0[/itex], we have: [itex]\psi(x,t) \equiv \int dx_0 G(x,t,x_0, t_0) \psi(x_0, t_0)[/itex]
    2. If [itex]t > t_1 > t_0[/itex], then [itex]G(x,t,x_0,t_0) = \int dx_1 G(x,t, x_1, t_1) G(x_1, t_1, x_0, t_0)[/itex]
    I don't think that the existence of such a Green's function depends on the Hamiltonian being time-independent, although the nice representation in terms of energy eigenfunctions is only possible then.
     
  6. Feb 6, 2017 #5

    ShayanJ

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    Yeah, its true that the existence of the Green function doesn't depend on the Hamiltonian being time-independent, but how can you prove ## \displaystyle \Psi(x,t) \propto \int_{z(t=-\infty)=const}^{z(t)=x} \mathcal D[z(t)] \ e^{iS[z(t)]} ## using that Green function?
     
  7. Feb 7, 2017 #6

    samalkhaiat

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    [tex]|\psi (t)\rangle = G(t,t_{0}) |\psi (t_{0}) \rangle .[/tex] In coordinate space this becomes [tex]\psi (x,t) = \int dy \ G(x,t ; y,t_{0}) \langle y | \psi (t_{0}) \rangle .[/tex] Suppose that at time [itex]t_{0}[/itex], the particle is at [itex]x_{0}[/itex], i.e., [itex]|\psi (t_{0})\rangle = |x_{0}\rangle[/itex]. Then [tex]\psi (x,t) = \int dy \ G(x,t ; y,t_{0}) \delta (y - x_{0}) = G(x,t ; x_{0},t_{0}) .[/tex] Now, a path integral expression for the Green’s function can be derived even when the Hamiltonian is time dependent: Instead of starting with [tex]G(t,t_{0}) = \theta(t-t_{0}) \ e^{- i (t-t_{0})H} ,[/tex] which is no longer true for [itex]H = H(t)[/itex], you can consider the following time-ordered product [tex]G(t,t_{0}) = \lim_{\epsilon \to 0} e^{-i\epsilon H(t)} e^{-i\epsilon H(t - \epsilon)} \ \cdots \ e^{-i\epsilon H(t_{0} + 2\epsilon)} e^{-i\epsilon H(t_{0} + \epsilon)} \equiv \mbox{T} \left( e^{-i \int_{t_{0}}^{t} dt \ H(t) } \right) ,[/tex] with [itex](t-t_{0})/ \epsilon[/itex] factors, as starting point for path integral formulation.
     
    Last edited: Feb 7, 2017
  8. Feb 8, 2017 #7

    ShayanJ

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    We have ## |\Psi(t)\rangle=G(t)|\Psi\rangle ##, so ## \langle x|\Psi(t)\rangle=\langle x,t|\Psi\rangle=\langle x|G(t)|\Psi\rangle ##, which means ## |x,t\rangle=G^\dagger(t) |x\rangle ##. Assuming that ## G^\dagger(t)=\mbox T \left( e^{i\int_0^t dt' \ H(t')} \right) ##, we'll have:
    ## \langle x|y,t \rangle=\langle x | \mbox T \left( e^{i\int_0^t dt' \ H(t')} \right) |y\rangle=\sum_b \langle x | \mbox T \left( e^{i\int_0^t dt' \ H(t')} \right) |\psi_b(t) \rangle \langle \psi_b(t) |y\rangle ##
    Where ## \{|\psi_b(t)\rangle \} ## is the complete set of energy eigenstates at time t.
    I don't see how to continue from there.
     
  9. Feb 8, 2017 #8

    stevendaryl

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    Expanding in terms of energy eigenfunctions is not necessary.

    The defining property of [itex]G[/itex] (the function, not the operator) is, as I said, that if [itex]\psi(x,t)[/itex] is a solution to Schrodinger's equation, then [itex]\psi(x,t) = \int dx_0 G(x,t, x_0, t_0) \psi(x_0, t_0)[/itex]. [itex]G(x,t,x_0, t_0)[/itex] has a kind of self-referential property:

    [itex]G(x,t,x_0, t_0) = \int dx_1 G(x,t, x_1, t_1) G(x_1, t_1, x_0, t_0)[/itex]

    (where [itex]t > t_1 > t_0[/itex])

    So you can keep expanding, so that for any [itex]N > 0[/itex],

    [itex]G(x,t,x_0,t_0) = \int dx_1 dx_2 ... dx_N \Pi_{j=0}^{N} G(x_{j+1}, t_0 + (j+1) \delta t, x_j, t_0 + j \delta t)[/itex]
    where [itex]\delta t = \frac{t-t_0}{N+1}[/itex] and [itex]x_{N+1} = x[/itex].

    This is an exact equality. But to go further toward the path integral, we need an approximation for [itex]G(x, t+\delta t, x', t)[/itex] for when [itex]\delta t[/itex] is small. I'm going to number the steps from here:

    1. Express [itex]G(x, t+\delta t, x', t)[/itex] in terms of the delta function.

    So here is an expression for [itex]G[/itex] for small [itex]\delta t[/itex]:

    [itex]G(x, t+\delta t, x', t) \approx e^{-i H(x, p, t) \delta t} \delta(x-x')[/itex]

    If [itex]H[/itex] were time-independent, then this would be valid for all [itex]\delta t[/itex], not just when [itex]\delta t[/itex] is small.

    2. Use the delta function to get rid of the x-dependence of the Hamiltonian

    Now, the next approximation is that since [itex]\delta(x-x')[/itex] is zero away from [itex]x=x'[/itex], we can replace [itex]x[/itex] by [itex]x'[/itex] in the exponential:

    [itex]e^{-i H(x, p, t) \delta t} \delta(x-x') \Rightarrow e^{-i H(x', p, t) \delta t} \delta(x-x')[/itex]

    3. Use the integral representation of the delta function.

    The next trick is to use the integral representation of the delta-function:

    [itex]\delta(x-x') = \frac{1}{2\pi} \int dk e^{i k (x-x')}[/itex]

    So we need to evaluate:

    [itex]\frac{1}{2\pi} e^{-i H(x',p, t) \delta t} \int dk e^{i k (x-x')}[/itex]

    We can bring the operator [itex]e^{-i H(x',p,t) \delta t}[/itex] inside the integral:

    [itex]\frac{1}{2\pi}\int dk e^{-i H(x',p, t) \delta t} e^{i k (x-x')}[/itex]

    4. Replace the operator [itex]p[/itex] by its eigenvalue.

    Since [itex]e^{i k (x-x')}[/itex] is an eigenstate of momentum, we can replace [itex]p[/itex] by its eigenvalue, [itex]k[/itex]. This gives us:

    [itex]\frac{1}{2\pi}\int dk e^{-i H(x',k,t) \delta t} e^{i k (x-x')} = \frac{1}{2\pi}\int dk e^{i (k \frac{x-x'}{\delta t} - H(x',k,t)) \delta t}[/itex]

    This is now a pure function expression, [itex]H(x',k,t)[/itex] is only a function, not an operator.

    5. Use method of stationary phase to evaluate the integral.

    Let's expand the expression in the exponential in a power series, centered on [itex]k=k_0[/itex] (we'll decide [itex]k_0[/itex] a little later):

    [itex]k \frac{x-x'}{\delta t} - H(x',k,t) \approx (k_0 \frac{x-x'}{\delta t} - H(x', k_0, t)) + (k-k_0) (\frac{x-x'}{\delta t} - \frac{\partial H(x,p,t)}{\partial p}|_{x=x', p=k_0}) + \frac{(k-k_0)^2}{2} \frac{\partial^2 H(x,p,t)}{\partial p^2}|_{x=x', p=k_0} + ...[/itex]

    The method of stationary phase chooses [itex]k_0[/itex] so that the linear term, proportional to [itex]k-k_0[/itex], vanishes. This means choosing [itex]k_0[/itex] so that

    [itex]\frac{x-x'}{\delta t} - \frac{\partial H(x,p,t)}{\partial p}|_{x=x', p=k_0} = 0[/itex]

    6. Use classical Hamiltonian equations to rewrite the expression in terms of velocity and Lagrangian

    Here's where the first miracle happens: The classical (non-quantum) Hamilton equations of motion are:

    [itex]\frac{dx}{dt} = \frac{\partial H}{\partial p}[/itex]
    [itex]\frac{dp}{dt} = - \frac{\partial H}{\partial x}[/itex]

    So the stationary phase condition is just that [itex]\frac{x-x'}{\delta t} = v[/itex], where [itex]v[/itex] is the classical velocity associated with the Hamiltonian [itex]H(x,p,t)[/itex] when[itex]x=x', p=k_0[/itex].

    The quadratic term is proportional to [itex]\frac{\partial^2 H(x,p,t)}{\partial p^2}|_{x=x', p=k_0}[/itex], which can be written as [itex]\frac{\partial v}{\partial p}|_{p=k_0, v = \frac{x-x'}{\delta t}}[/itex]. For a simple nonrelativistic Hamiltonian of the form [itex]\frac{p^2}{2m} + V(x,t)[/itex], [itex]v = \frac{p}{m}[/itex], so [itex]\frac{\partial v}{\partial p} = \frac{1}{m}[/itex]. I'm going to assume that that's the case.

    Here's where the second miracle happens: The constant term in the Taylor expansion is:

    [itex]k_0 \frac{x-x'}{\delta t} - H(x', k_0, t)[/itex]

    For our choice of [itex]k_0[/itex], this is the same as

    [itex](p v - H(x,p,t))|_{p = k_0, v = \frac{x-x'}{\delta t}, x = x'}[/itex]

    That's the classical expression for the Lagrangian [itex]L(x,v,t)[/itex] in terms of the Hamiltonian [itex]H(x,p,t)[/itex]. So putting everything together, our exponent can be written as

    [itex][ L(x,v,t)|_{x=x', v= \frac{x-x'}{\delta t}} - \frac{1}{2m} (k-k_0)^2 + ...] i \delta t[/itex]

    So we can write (keeping only up to the quadratic term):

    [itex]G(x,t+\delta t, x', t) \approx \frac{1}{2m} e^{i L(x,v,t) \delta t} \int dk e^{-i \frac{1}{2m} (k-k_0)^2 \delta t}[/itex]

    We still have an integral, and it's actually divergent, but if we pretend that [itex]\delta t[/itex] has a small negative imaginary component, then we can evaluate it as a Gaussian: [itex] \int dk e^{-i \frac{1}{2m} (k-k_0)^2 \delta t} = \sqrt{\frac{2\pi m}{i \delta t}}[/itex]. So our final expression for [itex]G[/itex]:

    [itex]G(x, t+\delta t, x', t) \approx \sqrt{\frac{m}{2 \pi i \delta t}} e^{i L(x',v,t) \delta t}[/itex]

    Putting this back into our expression for [itex]G(x, t, x_0, t_0)[/itex] gives:

    [itex]G(x,t,x_0, t_0) \approx (\sqrt{\frac{m}{2 \pi i \delta t}})^N \int dx_1 dx_2 ... dx_N \Pi_{j=0}^N e^{i L(x_j, v_j, t_j) \delta t}[/itex] (where [itex]v_j = \frac{x_{j+1} - x_j}{\delta t}[/itex] and where [itex]t_j = t + j \delta t[/itex])

    [itex] = (\sqrt{\frac{m}{2 \pi i \delta t}})^N \int dx_1 dx_2 ... dx_N e^{i \sum_{j=0}^N L(x_j, v_j, t_j) \delta t}[/itex]

    Since [itex]\psi(x,t) = \int dx_0 G(x,t, x_0, t_0) \psi(x_0, t_0)[/itex], we can immediately write an expression for [itex]\psi[/itex]:

    [itex]\psi(x,t) = (\sqrt{\frac{m}{2 \pi i \delta t}})^N \int dx_0 dx_1 dx_2 ... dx_N e^{i \sum_{j=0}^N L(x_j, v_j, t_j) \delta t} \psi(x_0, t_0) [/itex]

    To the extent that the limit [itex]N \Rightarrow \infty[/itex] makes sense, we can write this as:

    [itex]\psi(x,t) = \int \mathcal{D}[z(t)] e^{i \int_{x_0}^x L(z, \frac{dz}{dt}, t) dt} \psi(x_0, t_0)[/itex]
     
  10. Feb 9, 2017 #9

    ShayanJ

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    I really appreciate it @stevendaryl, but that's not the formula I'm talking about.
    The formula I want to prove, is ## \displaystyle \Psi(x,t)\propto \int_{z(t=-\infty)=const}^{z(t)=x} \mathcal D[z(t)] \ e^{iS[z(t)]} ##. Notice that there is no ## \Psi(const,-\infty) ## under the integral sign. Also the boundary conditions are at ## -\infty ## and t. Its interpretation is different too. Your formula is somehow ## |\Psi(t)\rangle=U(t,t_0)|\Psi(t_0)\rangle ## in path integral form but the formula I mentioned is a way of writing a wave-function if the system was at the ground state at ## -\infty ##. Its the generalization of the formula in post #3 for time-dependent Hamiltonians.
     
  11. Feb 9, 2017 #10

    stevendaryl

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    Sorry.

    But in the case of a time-dependent Hamiltonian, is there a notion of "ground state"? Or are you assuming some particular form for [itex]H[/itex] in the [itex]t \rightarrow -\infty[/itex] limit?
     
  12. Feb 9, 2017 #11

    ShayanJ

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    Well, its the same assumption people make when doing scattering calculations, that there is no interaction at ## t=\pm \infty ##. So any time-dependence in the Hamiltonian is only there for some finite period of time. Then we can ask what happens to the ground state of ## H(t=-\infty) ## if we allow it to evolve via ## H(t) ##.
     
  13. Feb 9, 2017 #12

    stevendaryl

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    I might be missing something, but isn't what you're asking for the same as

    [itex]lim_{t_0 \rightarrow -\infty} U(t,t_0) \Psi(x, t_0)[/itex]

    Why do you say the interpretation is different?
     
  14. Feb 9, 2017 #13

    ShayanJ

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    That's true, but the difference is your formula gave the wave-function at time ##t##, in terms of the wave-function at time ##t_0##. But the formula I wrote, doesn't have any wave-function under the integral sign. What I mean by a different interpretation, is that your formula takes a state at some time and gives another state at a later time. But my formula just gives the wave-function at time t, with the assumption that the wave-function at ## -\infty ## was the ground state wave-function.
     
  15. Feb 9, 2017 #14

    stevendaryl

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    I apologize for misunderstanding what you were asking, but let me ask another clarifying question.

    I did some Googling about "ground state" and "path integral", so maybe I know what you're talking about now.

    In the case of a time-independent Hamiltonian, you can use the energy eigenvalue representation of the Green's function to show that

    [itex]lim_{t_0 \rightarrow +i \infty} G(x,t, x_0, t_0) \propto \psi_0(x,t)[/itex]

    (Because the contribution of eigenstate [itex]j[/itex] has a factor [itex]e^{-i E_j (t-t_0)}[/itex], so taking the limit as [itex]t_0 \rightarrow i \infty[/itex] kills off all contributions except the one with lowest energy).

    So are you asking about the generalization of this to the case where [itex]H[/itex] is time-dependent? There isn't a notion of "ground state" in that case, is there? I suppose you could still, for a fixed value of [itex]t[/itex], get a complete set of eigenstates for [itex]H(t)[/itex], and you could consider the "ground state" at each moment (the state with the lowest eigenvalue), but there is no guarantee that the ground state at one moment would evolve into the ground state at a later moment.
     
  16. Feb 9, 2017 #15

    ShayanJ

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    Yes, that's what I'm asking.
    And you're right, its not very helpful to talk about the ground state in an arbitrary time. What I'm saying is that the time-dependent Hamiltonian has a ground state at ## t=-\infty ##. We can take that as the initial condition and ask what happens if it evolves via the time-dependent Hamiltonian of the system. I want to prove that such a state can be represented by the formula ## \displaystyle \Psi(x,t)\propto \int_{z(t=-\infty)=const}^{z(t)=x} \mathcal D[z(t)] \ e^{iS[z(t)]} ##.
     
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